Wikipedia:Reference desk/Archives/Mathematics/2015 May 5

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May 5

Dual of a trapezoid

What is the dual of a trapezoid?? What special properties will this kind of quadrilateral always have?? Georgia guy (talk) 00:40, 5 May 2015 (UTC)

I think it depends on the context and how you're defining a dual. For regular polygons such subtleties aren't important because you get the same answer, at least up to similarity. For general polygons you get different answers depending on which "type" of dual you're talking about. If the context is projective geometry, which is where you usually talk about duality, then a trapezoid is just like any other quadrilateral and the question doesn't really make sense. Our article Dual polygon talks about rectification and polar duality as two possible definitions. For rectification, I suppose the dual of a polygon is defined by taking the midpoint of each side. (The article isn't that clear on the exact definition.) If this is taken as the definition then the dual of any quadrilateral is a parallelogram and any parallelogram is the dual of a trapezoid. So there is nothing to distinguish a dual of a trapezoid from the dual of any other quadrilateral. To me at least, the polar dual is the more natural definition but the concepts are more difficult and the definition still depends on some arbitrary choices. Basically, a polar dual is defined with respect to a given fixed conic. For a regular polygon the conic is implicitly taken as a circle with the same center; the radius doesn't really matter since you get the same dual up to similarity. For an arbitrary quadrilateral there isn't really a natural choice for which conic to use and you get different results for the dual depending on the choice, even if you restrict to circles. But assuming you have a fixed conic in mind, the polar dual of of a pair of parallel lines is a pair of points which are collinear with the center of the conic. So the polar dual of a trapezoid would be a quadrilateral where the center of the conic lies on one of the diagonals. Our article Dual polyhedron talks a bit more about polar duality than the one on polygons. --RDBury (talk) 06:10, 5 May 2015 (UTC)

Help meh wid muh calculus homework

Line integrals in multivariable calculus.

I'm beating my head against the wall over problem #5, photographed here. http://imgur.com/9JcmiAu

Parameter I chose was x=x and y=x^(1/2). Integral limits are from 1 to 4.) when you substitute the parameters back into the integral, you get the following:

Integrate (from 1 to 4): [x^2 * x^(3/2) - x^(1/2)] times SOMETHING dx.

Then I look at the solutions book photographed here: http://imgur.com/gcrkOHS

It tells me I got the parameters right, and that I set up the integral correctly up until the SOMETHING part. In the photo I circled the part that I'm not getting.

I was taught that after substituting the reparameterized functions back into the integral, you multiply everything by radical((dx/dt)^2+(dy/dt)^2)). In this case t is x so you get (dx/dx)^2 is 1 of course and (dy/dx)^2 would be[ 1/([2x^(1/2))]^2 which would be 1/(4x), no? So the SOMETHING I'm gettings is radical (1 + 1/(4x)).

But according to the solutions manual, the SOMETHING I should be getting is simply (1/2x^(1/2)) which is just the derivative of x^(1/2).

What happened to radical ((dx/dt)^2+(dy/dt)^2))?

I am feeling very dumb.--Captain Breakfast (talk) 18:16, 5 May 2015 (UTC)

There's a difference between taking $\int f(x,y)ds\to \int f\left((x(t),y(t)\right)dt$ and what the answer key is showing you here, which is using a different type of substitution/parameterization. Let $y={\sqrt {x}}$, then $dy={\frac {1}{2{\sqrt {x}}}}dx$ as you noted. The point is, your "rule" of "multiply everything..." is incorrect, that only works for one class of substitutions for solving line integrals by parameterizing the curve f(x,y) by t->(x(t),y(t)), and where the differential is ds=arc length. Note especially that the differential is dy for this problem, whereas it is the more common ds for #1-4. Hope that helps, I'm very rusty teaching line integrals :) SemanticMantis (talk) 18:49, 5 May 2015 (UTC)

That helps a lot! I'm still a little fuzzy about the different types of substitutions/parameterization. I read the chapter and my notes several times but I still have trouble recognizing when to use what method. For example, why does ds not equal arc length in the problem referenced above.--Captain Breakfast (talk) 18:57, 5 May 2015 (UTC)

Actually, I re-read your answer and the dy part should have been the giveaway that I should use the derivative of y as the differential rather than the formula for arc length (ds). I think I understand now. I was wondering about that dy when I first saw the problem. Thanks.--Captain Breakfast (talk) 18:59, 5 May 2015 (UTC)

Great! There's a reason your instructor (probably) encouraged you to never drop the differential symbols when writing up your homework, this is is a good example of why. When teaching calc I, I often put in a problem like $\int _{0}^{1}3x^{2}+2x+5\;dt$ just to see who was paying attention :) SemanticMantis (talk) 20:32, 5 May 2015 (UTC)

Diabolical!--Captain Breakfast (talk) 05:18, 6 May 2015 (UTC)